What is the probability mcc will obtain fewer than 20 new clients(to 2 decimals)?

Given that West Virginia has one of the highest divorce rates in the nation with an annual rate of approximately 5 divorces per 1000 people (Centers for Disease Control and Prevention website, January 12, 2012). The Marital Counseling Center, Inc. (MCC) thinks that the high divorce rate in the state may require them to hire additional staff. Working with a consultant, the management of MCC has developed the following probability distribution for x = the number of new clients for marriage counseling for the next year.



Part A:

Is this probability distribution valid?

The sum of f(x) = 0.05 + 0.10 + 0.10 + 0.20 + 0.35 + 0.20 = 1

Since 0 ≤ f(x) ≤ 1, thus the distribution is a valid probability distribution.

Part B:

What is the probability MCC will obtain more than 30 new clients (to 2 decimals)?

The probability that the MCC will obtain more than 30 new clients is given by



Part C:

What is the probability MCC will obtain fewer than 20 new clients(to 2 decimals)?

The probability that MCC will obtain fewer than 20 new clients is given by



Part D

Compute the expected value

The expected value is given by



Part 5:

The variance is given by

  b.   the average rate of change of f(x) is less than the average rate of change of g(x).

step-by-step explanation:

since the length of the interval is the same in each case, we only need to compare the change in y-value to see which is greater.

f(-2) = -3f(1) = 6 . . . estimated from f(x) = (x+2)^2-3, which matches points on the graph (-1,-2) and (0, 1)

so the change in y-value for function f is (6 )) = 9.

the table tells us

g(-2) = 1g(1) = 31

so the change in y-value for function g is (31 -1) = 30.

the change in y-value of f(x) on the interval is less than the change in y-value of g(x) on the same interval, so the average rate of change of f(x) is less than the average rate of change of g(x).